Let $H_1:=(H_1,B_1)$ and $H_2:=(H_2,B_2)$ are two pre-Hilbert spaces. Remembering that 'pre-Hilbert' means that $B_1$ and $B_2$ are non-negative sesquilinear forms in $H_1$ and $H_2$, respectively. Thus, $B_1$ and $B_2$ do not define a norm (define only seminorms). I want to prove that: $H_1$ and $H_2$ are Hausdorff space if and only if $H_1 \times H_2$ is Hausdorff space.
It's clear that $H:=(H_1\times H_2, B)$ is pre-Hibert space, where $B=B_1+B_2$. Recalling that the topology in a 'pre-Hilbert' space is the topology generated by the seminorm (which is the sesquilinear form). I'm not able to taste the rest.
the answer seems long, but it is not, it is just detailed!
I assume that $B((x,y),(x',y')):=B_1(x,x')+B_2(y,y')$. So in the product space $H_1\times H_2$ with the topology induced by the seminorm induced by $B$,the basis consists of the sets $$D_{(x,y)}(r):=\{(z,w)\in H_1\times H_2: \sqrt{B_1(x-z,x-z)+B_2(y-w,y-w)}<r\} $$ with $(x,y)\in H_1\times H_2$ and $r>0$.
Step 1: Observe that the topology of $(H_1\times H_2, B)$ is equivalent to the product topology of $(H_1, B_1)$ and $(H_2,B_2)$.
The product topology has as a basis the sets $$D_x(r)\times D_y(r')=\{(z,w)\in H_1\times H_2: \sqrt{B_1(x-z,x-z)}<r\text{ and }\sqrt{B_2(y-w,y-w)}<r'\} $$ where $x\in H_1, y\in H_2$ and $r,r'>0$.
With some simple calculations we get that $D_{(x,y)}(r)\subset D_x(r)\times D_y(r)$ and that $D_x(r)\times D_y(r')\subset D_{(x,y)}(r+r')$, since for $s,t\geq0$ it is $\sqrt{s+t}\leq\sqrt{s}+\sqrt{t}$. This justifies that the topologies are equivalent.
Step 2: If $(X,\tau)$ and $(X,\tau')$ are two equivalent topologies on a space, then $(X,\tau)$ is Hausdorff if and only if $(X,\tau')$ is Hausdorff.
I leave this to you, it is easy.
Step 3: Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be two topological spaces. It is true that $(X\times Y,\tau)$ (where $\tau$ is the product topology) is Hausdorff if and only if both $(X,\tau_X)$ and $(Y,\tau_Y)$ are Hausdorff.
This is again easy: for the converse, suppose that $(x,y), (x',y')$ are distinct points in $X\times Y$. Then either $x\neq x'$ or $y\neq y'$. WLOG assume that $x\neq x'$. Since $X$ is Hausdorff we can find $U,V\subset X$ neighborhoods of $x,x'$ respectively with $U\cap V=\emptyset$. The sets $U\times Y$ and $V\times Y$ are disjoint neighborhoods of $(x,y)$ and $(x',y')$ respectively.
Conversely, if $X\times Y$ is Hausdorff and $x\neq x'$ take $y\in Y$ and consider $(x,y), (x',y)$. These are distinct points in $X\times Y$, which is Hausdorff, so take disjoint neighborhoods of these points. Apply the projection onto $X$ (which is an open map) to obtain disjoint neighborhoods of $x$ and $x'$, which shows $X$ is Hausdorff (likewise for $Y$).